f is not continuous at any point c of its domain R when the metric d on R is Euclidean

Question

Let $f\colon (\mathbb{R},d) \to (\mathbb{R}, |\cdot| )$ be defined as

$$f(x) = \begin{cases}\phantom{-}1, &x ∈ \mathbb{Q} \\-1, &x ∉ \mathbb{Q} \end{cases}$$

I want to prove that $f$ IS NOT continuous at any point $c$ of its domain $\mathbb{R}$ when the metric $d$ on $\mathbb{R}$ is Euclidean .

I want to prove that $f$ IS continuous at all points of its domain when the metric $d$ is the discrete metric.

I know: A function $f$ is called continuous at $x \in X$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $ρ(f(y), f(x)) < \varepsilon$ whenever $d(y,x) < \delta$.

AND

I know alternatively: $f\colon X\to Y$ is continuous iff whenever $x_n \to x$ in $X$ we have $f(x_n) \to f(x)$

Answer 1

Your second characterisation of continuity (called sequentially continuous) is the practical one in this case (and in the case of metric spaces, equivalent to continuity).

1) You want to prove that for a given $c\in \mathbb{R}$ the map $f$ is not continous at $c$. Let's assume that $c\in \mathbb{Q}$, so $f(c)=1$. So you want to find a sequence $x_n$ in $\mathbb{R}$ which converges to $c$ in the euclidean metric such that $f(x_n)=-1$ for all $n$. This would show that $f$ is not continous at $c$. Such a sequence exists, $f(x_n)=-1$ means that $x_n \notin \mathbb{Q}$ . Can you find a sequence in $\mathbb{R} \setminus \mathbb{Q}$ which converges to $c\in \mathbb{Q}$? The case $c\notin \mathbb{Q}$ is analogous.

2) To show that $f$ is continous for $d=\text{discrete}$, you have to understand first, what $x_n \to c$ means in the discrete metric. A sequence $x_n$ converges to $c$ if $d(x_n,c) \to 0$. But the distance in the discrete metric of nonequal points is always $1$. What does that mean for your sequence?

Answer 2

$\Bbb Q$ in $\Bbb R$ and so is $\Bbb R -\Bbb Q$. Hence for every real number $r$ there exist sequences $(a_k) \subset \Bbb Q$ and $(b_k) \subset \Bbb R - \Bbb Q$ such that $a_k \to r$ and $(b_k)\to r$. But $f(a_k)\to 1$ and $f(b_k)\to-1$. Hence f is not continuous. This answers your first question.

Answer 3

Hint: Use your last characterization and the sequence

$$x_n = c + \frac{\sqrt{2}}{n} \qquad y_n=c+\frac{1}{n}$$

May be useful separate cases when $c$ is or not rational.